Prove that V is a subspace of R4

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SUMMARY

The discussion centers on proving that the set V is a subspace of R4 by verifying three key properties: the existence of the zero vector, closure under addition, and closure under scalar multiplication. The participants emphasize that V consists of vectors x in R4 such that Ax = 0, where A is a linear transformation. To establish closure, it is necessary to demonstrate that if x and y are in V, then A(x+y) = 0 and A(tx) = 0 for any scalar t.

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Homework Statement


Prove that V is a subspace of R4

Actual problem is attached


Homework Equations



- S contains a zero element
- for any x in V and y in V, x + y is in V
- for any x in V and scalar k, kx is in V

The Attempt at a Solution



Its obvious that V is a subset of R4. And I know I must prove that is it contains zero vector which is obvious. But how to I prove its closure, I can see that it is by looking but how exactly do I denote it?
 

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If x and y are in V, that means that Ax=0 and Ay=0. Can you prove that A(x+y)=0? And can you prove that A(tx)=0 for every t a scalar? To prove that V is a subspace you need to prove that for every x,y in V, x+y and tx are also in V, which is precisely what the above shows
 
Office_Shredder said:
If x and y are in V, that means that Ax=0 and Ay=0. Can you prove that A(x+y)=0? And can you prove that A(tx)=0 for every t a scalar? To prove that V is a subspace you need to prove that for every x,y in V, x+y and tx are also in V, which is precisely what the above shows

So do these using the 0 vector?
 
Where does the 0 vector come in here? You're supposed to be looking at x and y any vectors in V. So the only property you can use about them is that Ax and Ay are both 0
 
Your set V is all the vectors x in R4 such that Ax = 0. Clearly A0 = 0, but there are other vectors in V, and you need to check that addition is closed for these vectors and scalar multiplication is closed also.
 

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